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May 10, 2018

Mixture and Alligation Problems with Short-cut Methods

Hi friends, I am Anonymous. A job aspirant and one of the ardent readers of Gr8ambitionz. Here I am sharing some problems on Mixture and Alligation with shortcut methods. If you like it, I will share more. Happy Reading :)

P.S : Sorry for the bad handwriting :P

1. 729 liters of a mixture has milk and water in the ratio of 7:2. How much water should be added to the mixture so that the ratio may become 7:3 ?

Solution :

2. In what proportion must tea at Rs. 60 per kg be mixed tea at Rs. 88 per kg. So that the mixture may be worth Rs. 68 kg ?

3. In what ratio should water be added to a liquid costing Rs. 12 per litre, so as to make a profit of 25% by selling the diluted liquid at Rs. 13.75 per litre ?

Solution :

Shortcut Method :

4. How much water should be added to 20 liter of milk at Rs. 10 as liter so as to have a mixture worth Rs. 8 a liter ?

Solution :

Let the quantity of water to be added be X liters

Cost of 20 liters of milk at Rs. 10 a liter = 20 x 10 = Rs. 200

=> Quantity of the mixture = (20 + X) liters

=> Cost of mixture at Rs. 8 a liter = (20 + X) 8 = Rs (160 + 8X)

=> 160 + 8X = 200

=> 8X = 200 - 160 = 40

So, X = 40/8 = 5 liters

5. A tea producer blends two qualities of tea from two different gardens, one costing Rs. 27 per kg and the other Rs. 30 per kg in the ratio 5 : 3. He sells the blended tea at Rs. 30.25 per kg. Find his percentage profit ?

Solution :

Suppose he blended 5X kg tea of first quality and 3X kg tea of second quality

=> Cost of 5X kg of tea = 5X x 27 = Rs. 135 X

And the cost of 3X kg of tea = 3X x 30 = Rs. 90X

=> Cost of 8X kg of the mixed tea = 135X + 90X = Rs. 225X

and S.P of 8X kg of the mixed tea = 8X x 30.25 = Rs. 242X

=> Gain 8X kg of the mixed tea = 242X - 225X = Rs. 17X

=> Percent profit = (17X x 100) / 225X = 7(5/9) %

6. Two vessels contain milk and water respectively in the ratio 3:1 and 5:3. Find the ratio in which these are to be mixed to get a new mixture in which the ratio of milk to water is 2:1.

Solution :

In the first vessel milk = 3/4 of the mixture

In the second vessel milk = 5/8 of the mixture

It is required have a mixture in which milk is 2/3rd of new mixture.

Let 1st mixture : 2nd mixture be 1:X

So the volume of resulting mixture = 1 + X

(milk in the 1st mix.) + (milk in the 2nd mix.) = (milk in the resulting mix.)